Not Applicable
Not Applicable
(1) Field of the Invention
The invention disclosed here generally pertains to the fields of geometry and trigonometry, and commercial applications thereof.
(2) Description of the Related Information Disclosed Under 37 C.F.R. 1.97 and 1.98.
The Austin triangles are similar in some ways to the slide rule. The slide rule consist of graduated scales capable of relative movement which can do simple calculations mechanically. In ordinary slide rules the operations include multiplication, division and extraction of square roots. The Austin triangles are different in that it has no moving parts and is used directly with an object or the xe2x80x9cfootprintxe2x80x9d of an object to calculate the area of a circle squared, square root of a sphere, cube root of a sphere, and double the volume of a sphere and a cube The protractor is limited in accuracy and would be cumbersome to use compared to the Austin trianglesin rapid extraction of square roots and cube roots. The slide rule works the formulas however, it cannot find angles of degree. The slide rule and protractor are related to the new invention but is limited in solving specific problems solved by the new invention.
The protractor is an instrument used for drawing and measuring angles. The protractor has the shape of a half circle that is marked with degrees. The Austin triangles use angles to calculate measurements of drawings and objects,
In the most general terms the invention disclosed herein comprises an important concept of simplifying mathematical calculations.
The invention consists of a graduated scale on a vertex angle of isosceles triangles capable of rapid extraction of the square area of a circle, square root of a sphere, doubling the volume of a cube, cubing the volume of a sphere, doubling the volume of a sphere.
There are several classic problems in geometry (and applications thereof) that the invention disclosed herein, dubbed the xe2x80x9cAustin trianglesxe2x80x9d, can solve:
1. Square the area of the circle. In other words, converting the area of a circular surface to a square surface having the same area as the circle.
2. Square the area of the sphere. In other words, converting the area of a spherical structure to cuboidal structure having the same area as the sphere.
3. Finding the cubed root of the volume of a sphere.
4. Double the volume of a cube.
5. Double the volume of a sphere.
These problems can be solved with the use of an isosceles triangle and four new formulas that will solve each problem that was developed by the inventor of this instrument.
Problem 1. Finding the square root of an area of a circle; the new formula a=d*0.886.
Problem 2. Finding the square root of an area of a sphere, the new formula a=d*1.772.
Problem 3. Finding the cube root of the volume of a sphere; the new formula v=d*0.80585.
Problem 4. Finding the cubed root of a cube that equals, double the volume of a cube. The new formula v=d*1.25992105.
Problem 5. Finding the cubed root of a sphere equals, double the volume of a sphere. Use problem three to solve cube root of a sphere, then use problem four to double volume.
Problem 1. Is to find the side of a square that is congruent to the area of a circle squared. I have discovered using the isosceles with the vertex angle of 52.589 and the diameter of the circle being congruent to each of the two corresponding sides of the isosceles triangle will multiply the base by 0.886 to equal the square root of the area of the circle square.
Problem 2. After you have the squared root of the area of the circle it is possible to multiply the base by two and get the square root of the area of the sphere squared. Formula to use is a=d*0.886*2, or a=d*1.772. Now it is possible to square the area of a circle and sphere.
Problem 3. In the process of problem one I discovered it is possible to find an indirect measurement, if the angle of the vertex is given. You can find the cubed root of the volume of the sphere; the problem is to find the side of a cube that is congruent to the volume of a sphere cubed. Using the isosceles with the vertex angle of 47 degrees and the diameter of the sphere being congruent to each of the sides will multiply the base by 0.80585 to equal the cube root of the volume cubed.
Problem 4. Is a classic, the problem is to find the side of a cube that is twice the volume of a given cube. I have discovered by using an isosceles with the vertex angle of 78.09 and the side of a cube being congruent to each of the two corresponding sides of the isosceles triangle will multiply the base by 1.25992105 to equal the cubed root of a cube having the volume of the cube doubled. Now it is possible to double the volume of a sphere when you find the cube root using the formula v=d*0.80585, then use the formula v=a+a*1.25992